Physics Asked on February 7, 2021
I learned from reading nLab (https://ncatlab.org/nlab/show/anti+de+Sitter+spacetime) that the anti-de Sitter Spacetime of dimension $d$, $AdS_d$, is homeomorphic to $mathbb{R}^{d-1} times S^1$. I tried to use the first image in this link for inspiration, but still I have a hard time conceptually understanding why these two spaces are homeomorphic. Can anyone explain? Also, what is the dimension of the AdS provided in this first image?
Something that is sometimes not discussed in detail is that what physicists call anti-de Sitter space is not in fact the submanifold
$$sum_{i = 1}^{d-1} (x_i)^2 - (x_d)^ 2 - (x_0)^2 = -R^2$$
of $mathbb{R}^{d-1,2}$. Take $d=2$ for concreteness: the defining equation becomes
$$x_1^2 - x_2^2 - x_0^2 = -R^2,$$
where $x_0$ and $x_2$ are the timelike coordinates. This is a one-sheet hyperboloid with rotation symmetry around the $x_1$ axis, which is the spacelike axis; therefore, its time coordinate corresponds to moving around the hyperboloid, and is thus periodic! It has the topology of $mathbb{R} times S^1$.
This is not very physically reasonable (or useful), so it's not what we usually call anti-de Sitter spacetime, and it's not what the image in the link shows. To get something a bit more realistic, we take the universal cover of this spacetime. You can look up the definition of universal cover if you want, but it will basically be another spacetime that shares the local metric properties but not the global topology of our original spacetime. In less abstract terms, we keep the same metric and "unroll" the hyperboloid so that the time coordinate is no longer periodic. This spacetime has topology $mathbb{R}^d$, and is what's shown in the picture.
Answered by Javier on February 7, 2021
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